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Proof step 9 from proof union-eq-empty (previous step)

∅ = zh0 ∪ zh1, zh0 ⊥ zh1 ⊦ ¬ r ∈ zh0 ∧ ¬ r0 ∈ zh1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ zh = zh0 ↔ (∀ r. (r ∈ zh ↔ r ∈ zh0) ∧ (r ∈ zh → zh[r] = zh0[r]))
Proof step 11:
zh0 ⊥ zh1, ∀ r. (r ∈ ∅ ↔ r ∈ zh0 ∪ zh1) ∧ (r ∈ ∅ → (∅)[r] = (zh0 ∪ zh1)[r]) ⊦ ¬ r ∈ zh0 ∧ ¬ r0 ∈ zh1

Rule argument

spec: heap, inst: zheap name: ext seq: ⊦ zh = zh0 ↔ (∀ r. (r ∈ zh ↔ r ∈ zh0) ∧ (r ∈ zh → zh[r] = zh0[r])) subst: [zh0, zh] → [zh0 ∪ zh1, ∅] flags: <all paths>

Additional information

Used terms: zh0 ∪ zh1, ∅